Given a polyhedron $P$ with vertices $V_i[x_i,y_i,z_i]$, how can we determine if there exist an affine transformation which transforms the polyhedron into $P'$ with vertices $V'_i[x'_i,y'_i,z'_i]$, where all $x'_i,y'_i,z'_i$ are rational? I am especially interested in the case when $P$ is regular (Platonic), uniform (Archimedean), Catalan or Johnson but my question can be generalised for any polyhedron.
For example, let $P$ be a regular icosahedron. Its vertices are all permutations of $[0, \pm 1, \pm \phi]$ with $\phi = (1+\sqrt{5})/2$. I am sure there is no way how we can write another parametrisation of vertices so a) it stays regular and b) the vertices are rational. However, is there at least an affine transfomation which transforms this regular icosahedron into a shape whose vertices are indeed rational?
No.
Suppose that we have a set of distinct points $P_1,\ldots,P_n\in\mathbb{Q}^3$. Then if we have three non-coplanar vectors $v_1=P_i-P_j, v_2=P_{i'}-P_{j'}, v_3=P_{i''}-P_{j''}$, every other vector between one point and another will be in the rational span of $v_1,v_2,v_3$.
However, the property we've just described is invariant under non-degenerate affine transformations, so any affine transformation of such a set of points (including translations) will preserve the property that every difference between points is in the rational span of a fixed set of three such differences.
The contrapositive of this claim is that any set of points without this property cannot be parametrized rationally under any affine transformation.
Now, take the icosahedron with coordinates $(0,\pm 1,\pm\varphi)$ and cyclic rotations thereof. We can take $v_1=(1,\varphi,0)-(-1,\varphi,0)=(2,0,0)$, and likewise $v_2=(0,2,0), v_3=(0,0,2)$. Then, if the icosahedron was rational under some affine transformation, every difference between vertices of the icosahedron would have to lie in $\text{Span}_{\mathbb Q}(v_1,v_2,v_3) = \mathbb{Q}^3$. But one such difference is the vector $(0,0,2\varphi)$, so we have a contradiction.
You can make a very similar argument to rule out the dodecahedron, and the other three platonic solids have rational-coordinate embeddings without any affine transformation needed. For the Archimedean and Catalan solids, you'll likely be able to quickly rule out many of them by identifying a subset of the vertices (or some rational sums of the vertices) which form a dodecahedron or icosahedron.
The one Archimedean case where this method doesn't seem to immediately work is the snub cube, but I suspect well-chosen vectors can get you the desired conclusion with a similar argument.
Note that for every Platonic or Archimedean solid, one can turn rational coordinates for one into rational coordinates for the dual, since the vertices of the dual are just the average of the vertices on each face.